SOLUTION MANUAL Game Theory Basics 1st Edition By Bernhard von Stengel. Chapters 1 - 12 NEWEST VERSION 2025

SOLUTION MANUAL
Game Theory Basics 1st Edition
By Bernhard von Stengel. Chapters 1 - 12




1

,TABLE OF CONTENTS R R R




1 - Nim and Combinatorial Games
R R R R R




2 - Congestion Games
R R R




3 - Games in Strategic Form
R R R R R




4 - Game Trees with Perfect Information
R R R R R R




5 - Expected Utility
R R R




6 - Mixed Equilibrium
R R R




7 - Brouwer’s Fixed-Point Theorem
R R R R




8 - Zero-Sum Games
R R R




9 - Geometry of Equilibria in Bimatrix Games
R R R R R R R




10 - Game Trees with Imperfect Information
R R R R R R




11 - Bargaining
R R




12 - Correlated Equilibrium
R R R




2

,Game Theory Basics
R R




Solutions R to R Exercises
© R Bernhard Rvon RStengel R2022

Solution Rto RExercise R1.1

(a) Let R≤ Rbe Rdefined Rby R(1.7). R To Rshow Rthat R≤ Ris Rtransitive, Rconsider Rx, Ry, Rz Rwith Rx R ≤ Ry Rand
Ry R≤ Rz. RIf Rx R= Ry Rthen Rx R≤ Rz, Rand Rif Ry R= Rz Rthen Ralso Rx R≤ Rz. RSo Rthe Ronly Rcase Rleft Ris Rx

R< Ry Rand R y R < R z, Rwhich Rimplies R x R < R z Rbecause R< Ris Rtransitive, Rand Rhence R x R ≤ Rz.

Clearly, R≤ Ris Rreflexive Rbecause Rx R= Rx Rand Rtherefore Rx R ≤ Rx.
To Rshow Rthat R≤R R R Ris Rantisymmetric, Rconsider Rx Rand Ry Rwith Rx≤ R R R R Ry Rand ≤
Ry R R R R Rx. RIf Rwe

Rhad Rx R≠ Ry Rthen Rx R< Ry Rand Ry R< Rx, Rand Rby Rtransitivity Rx R< Rx Rwhich Rcontradicts R(1.38).

RHence Rx R = R y, Ras Rrequired. R This Rshows Rthat R ≤ Ris Ra Rpartial Rorder.

Finally, Rwe Rshow R(1.6), Rso Rwe Rhave Rto Rshow Rthat Rx R< Ry Rimplies Rx R≤R Ry Rand Rx R≠ Ry Rand
Rvice Rversa. RLet Rx R< Ry, Rwhich Rimplies Rx Ry Rby R(1.7). RIf Rwe Rhad Rx R= Ry Rthen Rx R< Rx,
≤
Rcontradicting R(1.38), Rso Rwe Ralso Rhave Rx R≠ Ry. R Conversely, Rx R R R y Rand Rx R≠ Ry Rimply Rby R(1.7)
≤
Rx R < R y R or R x R = R y R where Rthe Rsecond Rcase Ris Rexcluded, Rhence R x R < R y, Ras Rrequired.


(b) Consider Ra Rpartial Rorder R≤ and Rassume R(1.6) Ras Ra Rdefinition Rof R<. RTo Rshow Rthat R< Ris
Rtransitive, Rsuppose Rx R< Ry, Rthat Ris, Rx Ry Rand Rx R≠ Ry, Rand Ry R< Rz, Rthat Ris, Ry Rz Rand Ry R≠ Rz.
≤ ≤
RBecause R R R Ris Rtransitive, Rx R R R Rz. RIf Rwe Rhad Rx R= Rz Rthen Rx R R R R Ry Rand Ry R R R R Rx Rand
≤ ≤ ≤ ≤
Rhence Rx R= Ry Rby Rantisymmetry Rof R R R R, Rwhich Rcontradicts R x R ≠ R y, Rso Rwe Rhave R x R R R R z
≤ ≤
R and R x R ≠ R z, Rthat Ris,Rx R < R z R by R(1.6), Ras Rrequired.


Also, R< Ris Rirreflexive, Rbecause Rx R< Rx Rwould Rby Rdefinition Rmean Rx R R≤Rx Rand Rx R≠ Rx, Rbut
Rthe Rlatter Ris Rnot Rtrue.


Finally, Rwe Rshow R(1.7), Rso Rwe Rhave Rto Rshow Rthat Rx R ≤ Ry Rimplies Rx R< Ry Ror Rx R= Ry Rand
Rvice Rversa, Rgiven Rthat R< Ris Rdefined Rby R(1.6). RLet Rx R≤ Ry. RThen Rif Rx R= Ry, Rwe Rare Rdone,
Rotherwise Rx R≠ Ry Rand Rthen Rby Rdefinition Rx R< Ry. RHence, Rx R≤ Ry Rimplies Rx R< Ry Ror Rx R= Ry.

RConversely, Rsuppose Rx R < R y Ror Rx R= Ry. R If Rx R < R y Rthen Rx R ≤ Ry Rby R(1.6), Rand Rif Rx R=

Ry Rthen Rx R ≤ R y Rbecause R ≤ Ris Rreflexive. R This Rcompletes Rthe Rproof.



Solution Rto RExercise R1.2

(a) In R analysing R the R games R of R three R Nim R heaps R where R one R heap R has R size R one, R we R first
R look Rat Rsome Rexamples, Rand Rthen Ruse Rmathematical Rinduction Rto Rprove Rwhat Rwe Rconjecture

Rto Rbe Rthe Rlosing Rpositions. RA Rlosing Rposition Ris Rone Rwhere Revery Rmove Ris Rto Ra

Rwinning Rposition, Rbecause Rthen Rthe Ropponent Rwill Rwin. R The Rpoint Rof Rthis Rexercise Ris Rto

Rformulate Ra Rprecise Rstatement Rto Rbe Rproved, Rand Rthen Rto Rprove Rit.


First, Rif Rthere Rare Ronly Rtwo Rheaps Rrecall Rthat Rthey Rare Rlosing Rif Rand Ronly Rif Rthe Rheaps
Rare Rof Requal Rsize. R If Rthey Rare Rof Runequal Rsize, Rthen Rthe Rwinning Rmove Ris Rto Rreduce

RtheRlarger Rheap Rso Rthat Rboth Rheaps Rhave Requal Rsize.




3

, Consider Rthree Rheaps Rof Rsizes R1, Rm, Rn, Rwhere R≤1 R R ≤R R Rm R R R R Rn. RWe Robserve Rthe
Rfollowing: R1, R1, Rm Ris Rwinning, Rby Rmoving Rto R1, R1, R0. RSimilarly, R1, Rm, Rm Ris Rwinning, Rby
Rmoving Rto R0, Rm, Rm. RNext, R1, R2, R3 Ris Rlosing R(observed Rearlier Rin Rthe Rlecture), Rand

Rhence R1, R2, Rn Rfor Rn R4 Ris Rwinning. R1, R3, Rn Ris Rwinning Rfor Rany Rn R3 Rby Rmoving Rto R1, R3,
≥ ≥
R2. RFor R1, R4, R5, Rreducing Rany Rheap Rproduces Ra Rwinning Rposition, Rso Rthis Ris Rlosing.


The Rgeneral Rpattern Rfor Rthe Rlosing Rpositions Rthus Rseems Rto Rbe: R1, Rm, Rm+R1, Rfor Reven
Rnumbers Rm. R This Rincludes Ralso Rthe Rcase Rm R= R0, Rwhich Rwe Rcan Rtake Ras Rthe Rbase Rcase Rfor

Ran Rinduction. R We Rnow Rproceed Rto Rprove Rthis Rformally.


First Rwe Rshow Rthat Rif Rthe Rpositions Rof Rthe Rform R1, Rm, Rn Rwith R≤m R R R R R Rn Rare Rlosing
Rwhen Rm Ris Reven Rand Rn R= Rm R1, Rthen Rthese Rare Rthe Ronly Rlosing Rpositions Rbecause Rany
+
Rother Rposition R1, Rm, Rn R with Rm R R n R is Rwinning. R Namely, Rif Rm R = Rn R then Ra Rwinning
≤
Rmove RfromR1, Rm, Rm Ris Rto R0, Rm, R m, Rso Rwe Rcan Rassume Rm R< Rn. R If Rm Ris Reven Rthen Rn R> Rm R
+
R 1 R(otherwise Rwe Rwould Rbe Rin Rthe Rposition R1, Rm, Rm R R 1) Rand Rso Rthe Rwinning Rmove Ris
+ +
Rto R1, Rm, Rm R R 1. RIf Rm Ris Rodd Rthen Rthe Rwinning Rmove Ris Rto R1, Rm, R m R1, Rthe Rsame Ras
– −
Rposition R1, Rm R1, Rm R(this Rwould R also R be R a R winning R move R from R 1, Rm, Rm R so R there R the

R winning R move R is R not R unique).


Second, Rwe Rshow Rthat Rany Rmove Rfrom R1, Rm, Rm R+ R1 Rwith Reven Rm Ris Rto Ra Rwinning Rposition,
using Ras Rinductive Rhypothesis Rthat R1, RmJ, RmJ R+ R1 Rfor Reven RmJ Rand RmJ R< Rm Ris Ra Rlosing
R

Rposition. RThe Rmove Rto R0, Rm, Rm R+ R1 Rproduces Ra Rwinning Rposition Rwith Rcounter-move

Rto R0, Rm, Rm. RA Rmove Rto R1, R m , Rm R+ R1 Rfor Rm R< Rm Ris Rto Ra Rwinning Rposition Rwith Rthe
J J


Rcounter-move Rto R1, Rm , Rm R+ R1 Rif Rm Ris Reven Rand Rto R1, Rm , Rm R− R1 Rif Rm Ris Rodd. RA
J J J J J J


Rmove Rto R1, Rm, R m Ris Rto Ra Rwinning Rposition Rwith Rcounter-move Rto R0, Rm, Rm. RA Rmove Rto R1,

Rm, Rm Rwith R m R< R m Ris Ralso Rto Ra Rwinning Rposition Rwith Rthe Rcounter-move Rto R1, Rm R− R1,
J J J


Rm Rif R m Ris Rodd, Rand Rto R1, R m R 1, Rm Rif Rm Ris Reven R(in Rwhich Rcase Rm R 1 R< Rm Rbecause
J J J J J J


Rm Ris Reven). RThis Rconcludes Rthe Rinduction Rproof.
+ +
This Rresult Ris Rin Ragreement Rwith Rthe Rtheorem Ron RNim Rheap Rsizes Rrepresented Ras Rsums Rof
Rpowers Rof R2: R 1 R
0
∗ +∗ +∗R m R R n Ris Rlosing Rif Rand Ronly Rif, Rexcept Rfor R2 , Rthe Rpowers Rof R2
Rmaking Rup Rm Rand Rn Rcome Rin Rpairs. RSo Rthese Rmust Rbe Rthe Rsame Rpowers Rof R2, Rexcept Rfor R1
0
R= R2 , Rwhich Roccurs Rin Ronly Rm Ror Rn, Rwhere Rwe Rhave Rassumed Rthat Rn Ris Rthe Rlarger Rnumber,

Rso R1 RappearsRin R the R representation R of R n: R We R have R m R = R 2
a R R R R R R2b R R R R R R2c for R a R >
+ + + ··· a R ·R · ·R 2
≥b R
R b R > R c R > R R R R R R R R 1, Rso R m R is R even, R and, R with R the R same R a, Rb, Rc, R. R. R., R n R = R 2
R R c + + +···+ +
2 1 R = R m R R R R 1.
R Then

∗1 R+R∗R R R +R m
∗ R R≡R R∗R n R R R R R R 0. R The R following R is R an Rexample R using R the R bit R representation
R where

m R = R12 R(which Rdetermines Rthe Rbit Rpattern R1100, Rwhich Rof Rcourse Rdepends Ron Rm):

1 = 0001
12 = 1100
13 = 1101
Nim-sum 0 = 0000

(b) We Ruse R(a). RClearly, R1, R2, R3 Ris Rlosing Ras Rshown Rin R(1.2), Rand Rbecause Rthe RNim-sum Rof
Rthe Rbinary Rrepresentations R01, R10, R11 Ris R00. RExamples Rshow Rthat Rany Rother Rposition

Ris Rwinning. RThe Rthree Rnumbers Rare Rn, Rn R 1, Rn R R 2. RIf Rn Ris Reven Rthen Rreducing Rthe
+ +
Rheap Rof Rsize Rn R2 Rto R1 Rcreates Rthe Rposition Rn, Rn R 1, R1 Rwhich Ris Rlosing Ras Rshown Rin
+ +
R(a). RIf Rn Ris Rodd, Rthen Rn R 1 Ris Reven Rand Rn R R R2 R= R n R R R1 R R R1 Rso Rby Rthe Rsame
+ + ( + )+
Rargument, Ra Rwinning Rmove Ris Rto Rreduce Rthe RNim Rheap Rof Rsize Rn Rto R1 R(which Ronly

Rworks Rif Rn R > R1).



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